JEE Exam > JEE Notes > Mathematics (Maths) for JEE Main & Advanced > Circle Solved Examples
Circle Solved Examples | Mathematics (Maths) for JEE Main & Advanced PDF Download
| Download, print and study this document offline |
Please wait while the PDF view is loading
Page 1
Solved Examples on Circle
JEE Mains
Q1.A circle of radius 5 units touches both the axes and lies in first quadrant. If the circle makes one
complete roll on ?? -axis along the positive direction of ?? -axis, then its equation in the new position
is
(a) ?? ?? +?? ?? +???? ???? -???? ?? +?????? ?? ?? =??
(b) ?? ?? +?? ?? +???? ???? +???? ?? +?????? ?? ?? =??
(c) ?? ?? +?? ?? -???? ???? -???? ?? +?????? ?? ?? =??
(d) None of these
Ans : (d) The ?? -coordinate of the new position of the circle is 5+ circumferrence of the first circle =
5+10?? The ?? -coordinate is 5 and the radius is also 5 .
Hence, the equation of the circle in the new position is (?? -5-10?? )
2
+(?? -5)
2
=(5)
2
??? 2
+25+100?? 2
-10?? +100?? -20???? +?? 2
+25-10?? =25
??? 2
+?? 2
-20???? -10?? -10?? +100?? 2
+100?? +25=0
Q2. The abscissae of ?? and ?? are the roots of the equation ?? ?? +?? ???? -?? ?? =?? and their
ordinates are the roots of the equation ?? ?? +?? ???? -?? ?? =?? . The equation of the circle with ???? as
diameter is
(a) ?? ?? +?? ?? +?? ???? +?? ???? -?? ?? -?? ?? =??
(b) ?? ?? +?? ?? +?? ???? +???? -?? ?? -?? ?? =??
(c) ?? ?? +?? ?? +?? ???? +?? ???? +?? ?? +?? ?? =??
(d) None of these
Ans: (a) Let ?? 1
,?? 2
and ?? 1
,?? 2
be roots of ?? 2
+2???? -?? 2
=0 and ?? 2
+2???? -?? 2
=0 respectively.
Then, ?? 1
+?? 2
=-2?? ,?? 1
?? 2
=-?? 2
and ?? 1
+?? 2
=-2?? ,?? 1
?? 2
=-?? 2
The equation of the circle with ?? (?? 1
,?? 1
) and ?? (?? 2
,?? 2
) as the end points of diameter is
(?? -?? 1
)(?? -?? 2
)+(?? -?? 1
)(?? -?? 2
)=0
?? 2
+?? 2
-?? (?? 1
+?? 2
)-?? (?? 1
+?? 2
)+?? 1
?? 2
+?? 1
?? 2
=0; ?? 2
+?? 2
+2???? +2???? -?? 2
-?? 2
=0
Q3. Let ?? (?? ,?? )=?? be the equation of a circle. If ?? (?? ,?? )=?? has equal roots ?? =?? ,?? and
?? (?? ,?? )=?? has roots ?? =
?? ?? ,?? , then the centre of the circle is
(a) (?? ,
????
????
)
(b) (
????
????
,?? )
(c) (-?? ,
????
????
)
(d) None of these
Page 2
Solved Examples on Circle
JEE Mains
Q1.A circle of radius 5 units touches both the axes and lies in first quadrant. If the circle makes one
complete roll on ?? -axis along the positive direction of ?? -axis, then its equation in the new position
is
(a) ?? ?? +?? ?? +???? ???? -???? ?? +?????? ?? ?? =??
(b) ?? ?? +?? ?? +???? ???? +???? ?? +?????? ?? ?? =??
(c) ?? ?? +?? ?? -???? ???? -???? ?? +?????? ?? ?? =??
(d) None of these
Ans : (d) The ?? -coordinate of the new position of the circle is 5+ circumferrence of the first circle =
5+10?? The ?? -coordinate is 5 and the radius is also 5 .
Hence, the equation of the circle in the new position is (?? -5-10?? )
2
+(?? -5)
2
=(5)
2
??? 2
+25+100?? 2
-10?? +100?? -20???? +?? 2
+25-10?? =25
??? 2
+?? 2
-20???? -10?? -10?? +100?? 2
+100?? +25=0
Q2. The abscissae of ?? and ?? are the roots of the equation ?? ?? +?? ???? -?? ?? =?? and their
ordinates are the roots of the equation ?? ?? +?? ???? -?? ?? =?? . The equation of the circle with ???? as
diameter is
(a) ?? ?? +?? ?? +?? ???? +?? ???? -?? ?? -?? ?? =??
(b) ?? ?? +?? ?? +?? ???? +???? -?? ?? -?? ?? =??
(c) ?? ?? +?? ?? +?? ???? +?? ???? +?? ?? +?? ?? =??
(d) None of these
Ans: (a) Let ?? 1
,?? 2
and ?? 1
,?? 2
be roots of ?? 2
+2???? -?? 2
=0 and ?? 2
+2???? -?? 2
=0 respectively.
Then, ?? 1
+?? 2
=-2?? ,?? 1
?? 2
=-?? 2
and ?? 1
+?? 2
=-2?? ,?? 1
?? 2
=-?? 2
The equation of the circle with ?? (?? 1
,?? 1
) and ?? (?? 2
,?? 2
) as the end points of diameter is
(?? -?? 1
)(?? -?? 2
)+(?? -?? 1
)(?? -?? 2
)=0
?? 2
+?? 2
-?? (?? 1
+?? 2
)-?? (?? 1
+?? 2
)+?? 1
?? 2
+?? 1
?? 2
=0; ?? 2
+?? 2
+2???? +2???? -?? 2
-?? 2
=0
Q3. Let ?? (?? ,?? )=?? be the equation of a circle. If ?? (?? ,?? )=?? has equal roots ?? =?? ,?? and
?? (?? ,?? )=?? has roots ?? =
?? ?? ,?? , then the centre of the circle is
(a) (?? ,
????
????
)
(b) (
????
????
,?? )
(c) (-?? ,
????
????
)
(d) None of these
Ans : (b) ??? (?? ,?? )=?? 2
+?? 2
+2???? +2???? +?? =0
Now, ?? (0,?? )=?? 2
+2???? +?? =0 and its roots are 2, 2. ?2+2=-2?? ,2×2=?? , i.e. ?? =-2,?? =
4
?? (?? ,0)=?? 2
+2???? +?? =0, and its roots are
4
5
,5.
?
4
5
+5=-2?? ,
4
5
×5=?? , i.e., ?? =
-29
10
,?? =4. Hence, centre of the circle =(-?? ,-?? )=(
29
10
,2).
Q4. The range of values of a for which the point (?? ,?? ) is outside the circles ?? ?? +?? ?? +???? ?? =??
and ?? ?? +?? ?? -???? ?? +???? =?? is
(a) (-8,-?? )?(-?? ,?? )?(?? ,+8)
(b) (-?? ,-?? )
(c) (-8,-?? )?(-?? ,+8)
(d) None of these
Ans : (a) For circle, ?? 2
+?? 2
+10?? =0;
?? 2
+(4)
2
+10?? >0??? 2
+10?? +16>0?(?? +8)(?? +2)>0??? <-8 or ?? >-2
For circle, ?? 2
+?? 2
-12?? +20=0;?? 2
+(4)
2
-12?? +20>0??? 2
-12?? +36>0
?(?? -6)
2
>0??? ??? ~{6}
Taking common values from (i) and (ii), ?? ?(-8,-8)?(-2,6)?(6,+8).
Q5. If the straight line ?? =???? is outside the circle ?? ?? +?? ?? -???? ?? +???? =?? , then
(a) ?? >??
(b) ?? <??
(c) |?? |>??
(d) |?? |<??
Ans: (d) If the straight line ?? =???? is outside the given circle then perpendicular distance of line
from centre of circle > radius of circle
10
v1+?? 2
>v10 ? (1+?? 2
)<10??? 2
<9 ?|?? |<3
Q6. The equations of any tangents to the circle ?? ?? +?? ?? -?? ?? +?? ?? -?? =?? is
(a) ?? =?? (?? -?? )+?? v?? +?? ?? -??
(b) ?? =???? +?? v?? +?? ??
(c) ?? =???? +?? v?? +?? ?? -??
(d) None of these
Ans : (a) Equation of circle is (?? -1)
2
+(?? +2)
2
=3
2
.
As any tangent to ?? 2
+?? 2
=3
2
is given by ?? =???? +3v1+?? 2
Any tangent to the given circle will be ?? +2=?? (?? -1)+3v1+?? 2
??? =?? (?? -1)+
3v1+?? 2
-2
Page 3
Solved Examples on Circle
JEE Mains
Q1.A circle of radius 5 units touches both the axes and lies in first quadrant. If the circle makes one
complete roll on ?? -axis along the positive direction of ?? -axis, then its equation in the new position
is
(a) ?? ?? +?? ?? +???? ???? -???? ?? +?????? ?? ?? =??
(b) ?? ?? +?? ?? +???? ???? +???? ?? +?????? ?? ?? =??
(c) ?? ?? +?? ?? -???? ???? -???? ?? +?????? ?? ?? =??
(d) None of these
Ans : (d) The ?? -coordinate of the new position of the circle is 5+ circumferrence of the first circle =
5+10?? The ?? -coordinate is 5 and the radius is also 5 .
Hence, the equation of the circle in the new position is (?? -5-10?? )
2
+(?? -5)
2
=(5)
2
??? 2
+25+100?? 2
-10?? +100?? -20???? +?? 2
+25-10?? =25
??? 2
+?? 2
-20???? -10?? -10?? +100?? 2
+100?? +25=0
Q2. The abscissae of ?? and ?? are the roots of the equation ?? ?? +?? ???? -?? ?? =?? and their
ordinates are the roots of the equation ?? ?? +?? ???? -?? ?? =?? . The equation of the circle with ???? as
diameter is
(a) ?? ?? +?? ?? +?? ???? +?? ???? -?? ?? -?? ?? =??
(b) ?? ?? +?? ?? +?? ???? +???? -?? ?? -?? ?? =??
(c) ?? ?? +?? ?? +?? ???? +?? ???? +?? ?? +?? ?? =??
(d) None of these
Ans: (a) Let ?? 1
,?? 2
and ?? 1
,?? 2
be roots of ?? 2
+2???? -?? 2
=0 and ?? 2
+2???? -?? 2
=0 respectively.
Then, ?? 1
+?? 2
=-2?? ,?? 1
?? 2
=-?? 2
and ?? 1
+?? 2
=-2?? ,?? 1
?? 2
=-?? 2
The equation of the circle with ?? (?? 1
,?? 1
) and ?? (?? 2
,?? 2
) as the end points of diameter is
(?? -?? 1
)(?? -?? 2
)+(?? -?? 1
)(?? -?? 2
)=0
?? 2
+?? 2
-?? (?? 1
+?? 2
)-?? (?? 1
+?? 2
)+?? 1
?? 2
+?? 1
?? 2
=0; ?? 2
+?? 2
+2???? +2???? -?? 2
-?? 2
=0
Q3. Let ?? (?? ,?? )=?? be the equation of a circle. If ?? (?? ,?? )=?? has equal roots ?? =?? ,?? and
?? (?? ,?? )=?? has roots ?? =
?? ?? ,?? , then the centre of the circle is
(a) (?? ,
????
????
)
(b) (
????
????
,?? )
(c) (-?? ,
????
????
)
(d) None of these
Ans : (b) ??? (?? ,?? )=?? 2
+?? 2
+2???? +2???? +?? =0
Now, ?? (0,?? )=?? 2
+2???? +?? =0 and its roots are 2, 2. ?2+2=-2?? ,2×2=?? , i.e. ?? =-2,?? =
4
?? (?? ,0)=?? 2
+2???? +?? =0, and its roots are
4
5
,5.
?
4
5
+5=-2?? ,
4
5
×5=?? , i.e., ?? =
-29
10
,?? =4. Hence, centre of the circle =(-?? ,-?? )=(
29
10
,2).
Q4. The range of values of a for which the point (?? ,?? ) is outside the circles ?? ?? +?? ?? +???? ?? =??
and ?? ?? +?? ?? -???? ?? +???? =?? is
(a) (-8,-?? )?(-?? ,?? )?(?? ,+8)
(b) (-?? ,-?? )
(c) (-8,-?? )?(-?? ,+8)
(d) None of these
Ans : (a) For circle, ?? 2
+?? 2
+10?? =0;
?? 2
+(4)
2
+10?? >0??? 2
+10?? +16>0?(?? +8)(?? +2)>0??? <-8 or ?? >-2
For circle, ?? 2
+?? 2
-12?? +20=0;?? 2
+(4)
2
-12?? +20>0??? 2
-12?? +36>0
?(?? -6)
2
>0??? ??? ~{6}
Taking common values from (i) and (ii), ?? ?(-8,-8)?(-2,6)?(6,+8).
Q5. If the straight line ?? =???? is outside the circle ?? ?? +?? ?? -???? ?? +???? =?? , then
(a) ?? >??
(b) ?? <??
(c) |?? |>??
(d) |?? |<??
Ans: (d) If the straight line ?? =???? is outside the given circle then perpendicular distance of line
from centre of circle > radius of circle
10
v1+?? 2
>v10 ? (1+?? 2
)<10??? 2
<9 ?|?? |<3
Q6. The equations of any tangents to the circle ?? ?? +?? ?? -?? ?? +?? ?? -?? =?? is
(a) ?? =?? (?? -?? )+?? v?? +?? ?? -??
(b) ?? =???? +?? v?? +?? ??
(c) ?? =???? +?? v?? +?? ?? -??
(d) None of these
Ans : (a) Equation of circle is (?? -1)
2
+(?? +2)
2
=3
2
.
As any tangent to ?? 2
+?? 2
=3
2
is given by ?? =???? +3v1+?? 2
Any tangent to the given circle will be ?? +2=?? (?? -1)+3v1+?? 2
??? =?? (?? -1)+
3v1+?? 2
-2
Q7. If the distances from the origin to the centres of three circles ?? ?? +?? ?? +?? ?? ?? ?? -?? ?? =?? (?? =
?? ,?? ,?? ) are in G.P. then the lengths of the tangents drawn to them from any point on the circle
?? ?? +?? ?? =?? ?? are in
(a) A.P.
(b) G.P .
(c) H.P.
(d) None of these
Ans : (b) The centres of the given circles are (-?? ?? ,0)(?? =1,2,3)
The distances from the origin to the centres are ?? ?? (?? =1,2,3). It is given that ?? 2
2
=?? 1
?? 3
.
Let ?? (h,?? ) be any point on the circle ?? 2
+?? 2
=?? 2
, then,
h
2
+?? 2
=?? 2
Now, ?? ?? = length of the tangent from (h,?? ) to ?? 2
+?? 2
+2?? ?? ?? -?? 2
=0
=vh
2
+?? 2
+2?? ?? h-?? 2
=v?? 2
+2?? ?? h-?? 2
=v2?? ?? h
[?h
2
+?? 2
=?? 2
and ?? =1,2,3]
Therefore, ?? 2
2
=2?? 2
h=2h(v?? 1
?? 3
)
[??? 2
2
=?? 1
?? 3
]
=v2h?? 1
v2h?? 3
=?? 1
?? 3
. Hence, ?? 1
,?? 2
,?? 3
are in G.P .
Q8. The angle between a pair of tangents drawn from a point ?? to the circle ?? ?? +?? ?? +?? ?? -?? ?? +
?? ?????? ?? ?? +???? ?????? ?? ?? =?? is
(a) ?? ?? +?? ?? +?? ?? -?? ?? +?? =??
(b) ?? ?? +?? ?? +?? ?? -?? ?? -?? =??
(c) ?? ?? +?? ?? +?? ?? -?? ?? -?? =??
(d) ?? ?? +?? ?? +?? ?? -?? ?? +?? =??
Ans : (d) The centre of the circle ?? 2
+?? 2
+4?? -6?? +9sin
2
?? +13cos
2
?? =0 is ?? (-2,3) and its
radius is
v2
2
+(-3)
2
-9sin
2
?? -13cos
2
?? =v4+9-9sin
2
?? -13cos
2
?? =2sin ??
Let ?? (h,?? ) be any point on the locus. The ??????? =?? . Also ??????? =?? /2 i.e. triangle ?????? is a right
angle triangle.
Thus sin ?? =
????
????
=
2sin ?? v(h+2)
2
+(?? -3)
2
?v(h+2)
2
+(?? -3)
2
=2?(h+2)
2
+(?? -3)
2
=4or h
2
+?? 2
+4h-6?? +9=0
Thus the required equation of the locus is ?? 2
+?? 2
+4?? -6?? +9=0.
Page 4
Solved Examples on Circle
JEE Mains
Q1.A circle of radius 5 units touches both the axes and lies in first quadrant. If the circle makes one
complete roll on ?? -axis along the positive direction of ?? -axis, then its equation in the new position
is
(a) ?? ?? +?? ?? +???? ???? -???? ?? +?????? ?? ?? =??
(b) ?? ?? +?? ?? +???? ???? +???? ?? +?????? ?? ?? =??
(c) ?? ?? +?? ?? -???? ???? -???? ?? +?????? ?? ?? =??
(d) None of these
Ans : (d) The ?? -coordinate of the new position of the circle is 5+ circumferrence of the first circle =
5+10?? The ?? -coordinate is 5 and the radius is also 5 .
Hence, the equation of the circle in the new position is (?? -5-10?? )
2
+(?? -5)
2
=(5)
2
??? 2
+25+100?? 2
-10?? +100?? -20???? +?? 2
+25-10?? =25
??? 2
+?? 2
-20???? -10?? -10?? +100?? 2
+100?? +25=0
Q2. The abscissae of ?? and ?? are the roots of the equation ?? ?? +?? ???? -?? ?? =?? and their
ordinates are the roots of the equation ?? ?? +?? ???? -?? ?? =?? . The equation of the circle with ???? as
diameter is
(a) ?? ?? +?? ?? +?? ???? +?? ???? -?? ?? -?? ?? =??
(b) ?? ?? +?? ?? +?? ???? +???? -?? ?? -?? ?? =??
(c) ?? ?? +?? ?? +?? ???? +?? ???? +?? ?? +?? ?? =??
(d) None of these
Ans: (a) Let ?? 1
,?? 2
and ?? 1
,?? 2
be roots of ?? 2
+2???? -?? 2
=0 and ?? 2
+2???? -?? 2
=0 respectively.
Then, ?? 1
+?? 2
=-2?? ,?? 1
?? 2
=-?? 2
and ?? 1
+?? 2
=-2?? ,?? 1
?? 2
=-?? 2
The equation of the circle with ?? (?? 1
,?? 1
) and ?? (?? 2
,?? 2
) as the end points of diameter is
(?? -?? 1
)(?? -?? 2
)+(?? -?? 1
)(?? -?? 2
)=0
?? 2
+?? 2
-?? (?? 1
+?? 2
)-?? (?? 1
+?? 2
)+?? 1
?? 2
+?? 1
?? 2
=0; ?? 2
+?? 2
+2???? +2???? -?? 2
-?? 2
=0
Q3. Let ?? (?? ,?? )=?? be the equation of a circle. If ?? (?? ,?? )=?? has equal roots ?? =?? ,?? and
?? (?? ,?? )=?? has roots ?? =
?? ?? ,?? , then the centre of the circle is
(a) (?? ,
????
????
)
(b) (
????
????
,?? )
(c) (-?? ,
????
????
)
(d) None of these
Ans : (b) ??? (?? ,?? )=?? 2
+?? 2
+2???? +2???? +?? =0
Now, ?? (0,?? )=?? 2
+2???? +?? =0 and its roots are 2, 2. ?2+2=-2?? ,2×2=?? , i.e. ?? =-2,?? =
4
?? (?? ,0)=?? 2
+2???? +?? =0, and its roots are
4
5
,5.
?
4
5
+5=-2?? ,
4
5
×5=?? , i.e., ?? =
-29
10
,?? =4. Hence, centre of the circle =(-?? ,-?? )=(
29
10
,2).
Q4. The range of values of a for which the point (?? ,?? ) is outside the circles ?? ?? +?? ?? +???? ?? =??
and ?? ?? +?? ?? -???? ?? +???? =?? is
(a) (-8,-?? )?(-?? ,?? )?(?? ,+8)
(b) (-?? ,-?? )
(c) (-8,-?? )?(-?? ,+8)
(d) None of these
Ans : (a) For circle, ?? 2
+?? 2
+10?? =0;
?? 2
+(4)
2
+10?? >0??? 2
+10?? +16>0?(?? +8)(?? +2)>0??? <-8 or ?? >-2
For circle, ?? 2
+?? 2
-12?? +20=0;?? 2
+(4)
2
-12?? +20>0??? 2
-12?? +36>0
?(?? -6)
2
>0??? ??? ~{6}
Taking common values from (i) and (ii), ?? ?(-8,-8)?(-2,6)?(6,+8).
Q5. If the straight line ?? =???? is outside the circle ?? ?? +?? ?? -???? ?? +???? =?? , then
(a) ?? >??
(b) ?? <??
(c) |?? |>??
(d) |?? |<??
Ans: (d) If the straight line ?? =???? is outside the given circle then perpendicular distance of line
from centre of circle > radius of circle
10
v1+?? 2
>v10 ? (1+?? 2
)<10??? 2
<9 ?|?? |<3
Q6. The equations of any tangents to the circle ?? ?? +?? ?? -?? ?? +?? ?? -?? =?? is
(a) ?? =?? (?? -?? )+?? v?? +?? ?? -??
(b) ?? =???? +?? v?? +?? ??
(c) ?? =???? +?? v?? +?? ?? -??
(d) None of these
Ans : (a) Equation of circle is (?? -1)
2
+(?? +2)
2
=3
2
.
As any tangent to ?? 2
+?? 2
=3
2
is given by ?? =???? +3v1+?? 2
Any tangent to the given circle will be ?? +2=?? (?? -1)+3v1+?? 2
??? =?? (?? -1)+
3v1+?? 2
-2
Q7. If the distances from the origin to the centres of three circles ?? ?? +?? ?? +?? ?? ?? ?? -?? ?? =?? (?? =
?? ,?? ,?? ) are in G.P. then the lengths of the tangents drawn to them from any point on the circle
?? ?? +?? ?? =?? ?? are in
(a) A.P.
(b) G.P .
(c) H.P.
(d) None of these
Ans : (b) The centres of the given circles are (-?? ?? ,0)(?? =1,2,3)
The distances from the origin to the centres are ?? ?? (?? =1,2,3). It is given that ?? 2
2
=?? 1
?? 3
.
Let ?? (h,?? ) be any point on the circle ?? 2
+?? 2
=?? 2
, then,
h
2
+?? 2
=?? 2
Now, ?? ?? = length of the tangent from (h,?? ) to ?? 2
+?? 2
+2?? ?? ?? -?? 2
=0
=vh
2
+?? 2
+2?? ?? h-?? 2
=v?? 2
+2?? ?? h-?? 2
=v2?? ?? h
[?h
2
+?? 2
=?? 2
and ?? =1,2,3]
Therefore, ?? 2
2
=2?? 2
h=2h(v?? 1
?? 3
)
[??? 2
2
=?? 1
?? 3
]
=v2h?? 1
v2h?? 3
=?? 1
?? 3
. Hence, ?? 1
,?? 2
,?? 3
are in G.P .
Q8. The angle between a pair of tangents drawn from a point ?? to the circle ?? ?? +?? ?? +?? ?? -?? ?? +
?? ?????? ?? ?? +???? ?????? ?? ?? =?? is
(a) ?? ?? +?? ?? +?? ?? -?? ?? +?? =??
(b) ?? ?? +?? ?? +?? ?? -?? ?? -?? =??
(c) ?? ?? +?? ?? +?? ?? -?? ?? -?? =??
(d) ?? ?? +?? ?? +?? ?? -?? ?? +?? =??
Ans : (d) The centre of the circle ?? 2
+?? 2
+4?? -6?? +9sin
2
?? +13cos
2
?? =0 is ?? (-2,3) and its
radius is
v2
2
+(-3)
2
-9sin
2
?? -13cos
2
?? =v4+9-9sin
2
?? -13cos
2
?? =2sin ??
Let ?? (h,?? ) be any point on the locus. The ??????? =?? . Also ??????? =?? /2 i.e. triangle ?????? is a right
angle triangle.
Thus sin ?? =
????
????
=
2sin ?? v(h+2)
2
+(?? -3)
2
?v(h+2)
2
+(?? -3)
2
=2?(h+2)
2
+(?? -3)
2
=4or h
2
+?? 2
+4h-6?? +9=0
Thus the required equation of the locus is ?? 2
+?? 2
+4?? -6?? +9=0.
Q9. Tangents are drawn from any point on the circle ?? ?? +?? ?? =?? ?? to the circle ?? ?? +?? ?? =?? ?? . If
the chord of contact touches the circle ?? ?? +?? ?? =?? ?? ,?? >?? , then
(a) a, b, c are in A.P .
(b) a, b, c are in G.P .
(c) a, b, c are in H.P . (d) a, c, b are in G.P .
Ans : (b) Chord of contact of any point (?? cos ?? ,?? sin ?? ) on 1
st
circle with respect to 2
nd
circle is
???? cos ?? +???? sin ?? =?? 2
This chord touches the circle ?? 2
+?? 2
=?? 2
,
Hence, Radius = Perpendicular distance of chord from centre.
?? =
?? 2
?? vcos
2
?? +sin
2
?? ??? 2
=???? .Hence ?? ,?? ,?? are in G.P .
Q10. A foot of the normal from the point (?? ,?? ) to a circle is (?? ,?? ) and a diameter of the circle has
the equation ?? ?? -?? =?? . Then the equation of the circle is
(a) ?? ?? +?? ?? +?? ?? -?? =??
(b) ?? ?? +?? ?? -?? ?? -?? =??
(c) ?? ?? +?? ?? -?? ?? -?? =??
(d) None of these.
Ans : (b) The line joining (4,3) and (2,1) is also along a diameter. So, the centre is the intersection of
the diameters 2?? -?? =2 and ?? -3=(?? -4). Solving these, the centre =(1,0)
? Radius = Distance between (1,0) and (2,1)=v2.
? Equation of circle (?? -1)
2
+?? 2
=(v2)
2
??? 2
+?? 2
-2?? -1=0
Q11. The polar of the point (?? ,-
?? ?? ) with respect to circle (?? -?? )
?? +?? ?? =?? is
(a) ?? ?? -???? ?? +?? =??
(b) ?? ?? -?? -???? =??
(c) ???? ?? -?? -???? =??
(d) ?? -???? ?? -?? =??
Ans : (b)
The polar of the point (5,-
1
2
) is ?? ?? 1
+?? ?? 1
+?? (?? +?? 1
)+?? (?? +?? 1
)+?? =0
?5?? -
1
2
?? -2(?? +5)+0+0=0?3?? -
?? 2
-10=0?6?? -?? -20=0.
Q12. The equation of the circle which touches the circle ?? ?? +?? ?? -?? ?? +?? ?? +???? =?? externally
and to which the lines ?? ?? -?? ???? -?? ?? +?? ?? =?? are normals, is
(a) ?? ?? +?? ?? -?? ?? -?? ?? -?? =??
(b) ?? ?? +?? ?? +?? ?? +?? ?? +?? =??
(c) ?? ?? +?? ?? -?? ?? -?? ?? +?? =??
(d) ?? ?? +?? ?? -?? ?? -?? ?? +?? =??
Page 5
Solved Examples on Circle
JEE Mains
Q1.A circle of radius 5 units touches both the axes and lies in first quadrant. If the circle makes one
complete roll on ?? -axis along the positive direction of ?? -axis, then its equation in the new position
is
(a) ?? ?? +?? ?? +???? ???? -???? ?? +?????? ?? ?? =??
(b) ?? ?? +?? ?? +???? ???? +???? ?? +?????? ?? ?? =??
(c) ?? ?? +?? ?? -???? ???? -???? ?? +?????? ?? ?? =??
(d) None of these
Ans : (d) The ?? -coordinate of the new position of the circle is 5+ circumferrence of the first circle =
5+10?? The ?? -coordinate is 5 and the radius is also 5 .
Hence, the equation of the circle in the new position is (?? -5-10?? )
2
+(?? -5)
2
=(5)
2
??? 2
+25+100?? 2
-10?? +100?? -20???? +?? 2
+25-10?? =25
??? 2
+?? 2
-20???? -10?? -10?? +100?? 2
+100?? +25=0
Q2. The abscissae of ?? and ?? are the roots of the equation ?? ?? +?? ???? -?? ?? =?? and their
ordinates are the roots of the equation ?? ?? +?? ???? -?? ?? =?? . The equation of the circle with ???? as
diameter is
(a) ?? ?? +?? ?? +?? ???? +?? ???? -?? ?? -?? ?? =??
(b) ?? ?? +?? ?? +?? ???? +???? -?? ?? -?? ?? =??
(c) ?? ?? +?? ?? +?? ???? +?? ???? +?? ?? +?? ?? =??
(d) None of these
Ans: (a) Let ?? 1
,?? 2
and ?? 1
,?? 2
be roots of ?? 2
+2???? -?? 2
=0 and ?? 2
+2???? -?? 2
=0 respectively.
Then, ?? 1
+?? 2
=-2?? ,?? 1
?? 2
=-?? 2
and ?? 1
+?? 2
=-2?? ,?? 1
?? 2
=-?? 2
The equation of the circle with ?? (?? 1
,?? 1
) and ?? (?? 2
,?? 2
) as the end points of diameter is
(?? -?? 1
)(?? -?? 2
)+(?? -?? 1
)(?? -?? 2
)=0
?? 2
+?? 2
-?? (?? 1
+?? 2
)-?? (?? 1
+?? 2
)+?? 1
?? 2
+?? 1
?? 2
=0; ?? 2
+?? 2
+2???? +2???? -?? 2
-?? 2
=0
Q3. Let ?? (?? ,?? )=?? be the equation of a circle. If ?? (?? ,?? )=?? has equal roots ?? =?? ,?? and
?? (?? ,?? )=?? has roots ?? =
?? ?? ,?? , then the centre of the circle is
(a) (?? ,
????
????
)
(b) (
????
????
,?? )
(c) (-?? ,
????
????
)
(d) None of these
Ans : (b) ??? (?? ,?? )=?? 2
+?? 2
+2???? +2???? +?? =0
Now, ?? (0,?? )=?? 2
+2???? +?? =0 and its roots are 2, 2. ?2+2=-2?? ,2×2=?? , i.e. ?? =-2,?? =
4
?? (?? ,0)=?? 2
+2???? +?? =0, and its roots are
4
5
,5.
?
4
5
+5=-2?? ,
4
5
×5=?? , i.e., ?? =
-29
10
,?? =4. Hence, centre of the circle =(-?? ,-?? )=(
29
10
,2).
Q4. The range of values of a for which the point (?? ,?? ) is outside the circles ?? ?? +?? ?? +???? ?? =??
and ?? ?? +?? ?? -???? ?? +???? =?? is
(a) (-8,-?? )?(-?? ,?? )?(?? ,+8)
(b) (-?? ,-?? )
(c) (-8,-?? )?(-?? ,+8)
(d) None of these
Ans : (a) For circle, ?? 2
+?? 2
+10?? =0;
?? 2
+(4)
2
+10?? >0??? 2
+10?? +16>0?(?? +8)(?? +2)>0??? <-8 or ?? >-2
For circle, ?? 2
+?? 2
-12?? +20=0;?? 2
+(4)
2
-12?? +20>0??? 2
-12?? +36>0
?(?? -6)
2
>0??? ??? ~{6}
Taking common values from (i) and (ii), ?? ?(-8,-8)?(-2,6)?(6,+8).
Q5. If the straight line ?? =???? is outside the circle ?? ?? +?? ?? -???? ?? +???? =?? , then
(a) ?? >??
(b) ?? <??
(c) |?? |>??
(d) |?? |<??
Ans: (d) If the straight line ?? =???? is outside the given circle then perpendicular distance of line
from centre of circle > radius of circle
10
v1+?? 2
>v10 ? (1+?? 2
)<10??? 2
<9 ?|?? |<3
Q6. The equations of any tangents to the circle ?? ?? +?? ?? -?? ?? +?? ?? -?? =?? is
(a) ?? =?? (?? -?? )+?? v?? +?? ?? -??
(b) ?? =???? +?? v?? +?? ??
(c) ?? =???? +?? v?? +?? ?? -??
(d) None of these
Ans : (a) Equation of circle is (?? -1)
2
+(?? +2)
2
=3
2
.
As any tangent to ?? 2
+?? 2
=3
2
is given by ?? =???? +3v1+?? 2
Any tangent to the given circle will be ?? +2=?? (?? -1)+3v1+?? 2
??? =?? (?? -1)+
3v1+?? 2
-2
Q7. If the distances from the origin to the centres of three circles ?? ?? +?? ?? +?? ?? ?? ?? -?? ?? =?? (?? =
?? ,?? ,?? ) are in G.P. then the lengths of the tangents drawn to them from any point on the circle
?? ?? +?? ?? =?? ?? are in
(a) A.P.
(b) G.P .
(c) H.P.
(d) None of these
Ans : (b) The centres of the given circles are (-?? ?? ,0)(?? =1,2,3)
The distances from the origin to the centres are ?? ?? (?? =1,2,3). It is given that ?? 2
2
=?? 1
?? 3
.
Let ?? (h,?? ) be any point on the circle ?? 2
+?? 2
=?? 2
, then,
h
2
+?? 2
=?? 2
Now, ?? ?? = length of the tangent from (h,?? ) to ?? 2
+?? 2
+2?? ?? ?? -?? 2
=0
=vh
2
+?? 2
+2?? ?? h-?? 2
=v?? 2
+2?? ?? h-?? 2
=v2?? ?? h
[?h
2
+?? 2
=?? 2
and ?? =1,2,3]
Therefore, ?? 2
2
=2?? 2
h=2h(v?? 1
?? 3
)
[??? 2
2
=?? 1
?? 3
]
=v2h?? 1
v2h?? 3
=?? 1
?? 3
. Hence, ?? 1
,?? 2
,?? 3
are in G.P .
Q8. The angle between a pair of tangents drawn from a point ?? to the circle ?? ?? +?? ?? +?? ?? -?? ?? +
?? ?????? ?? ?? +???? ?????? ?? ?? =?? is
(a) ?? ?? +?? ?? +?? ?? -?? ?? +?? =??
(b) ?? ?? +?? ?? +?? ?? -?? ?? -?? =??
(c) ?? ?? +?? ?? +?? ?? -?? ?? -?? =??
(d) ?? ?? +?? ?? +?? ?? -?? ?? +?? =??
Ans : (d) The centre of the circle ?? 2
+?? 2
+4?? -6?? +9sin
2
?? +13cos
2
?? =0 is ?? (-2,3) and its
radius is
v2
2
+(-3)
2
-9sin
2
?? -13cos
2
?? =v4+9-9sin
2
?? -13cos
2
?? =2sin ??
Let ?? (h,?? ) be any point on the locus. The ??????? =?? . Also ??????? =?? /2 i.e. triangle ?????? is a right
angle triangle.
Thus sin ?? =
????
????
=
2sin ?? v(h+2)
2
+(?? -3)
2
?v(h+2)
2
+(?? -3)
2
=2?(h+2)
2
+(?? -3)
2
=4or h
2
+?? 2
+4h-6?? +9=0
Thus the required equation of the locus is ?? 2
+?? 2
+4?? -6?? +9=0.
Q9. Tangents are drawn from any point on the circle ?? ?? +?? ?? =?? ?? to the circle ?? ?? +?? ?? =?? ?? . If
the chord of contact touches the circle ?? ?? +?? ?? =?? ?? ,?? >?? , then
(a) a, b, c are in A.P .
(b) a, b, c are in G.P .
(c) a, b, c are in H.P . (d) a, c, b are in G.P .
Ans : (b) Chord of contact of any point (?? cos ?? ,?? sin ?? ) on 1
st
circle with respect to 2
nd
circle is
???? cos ?? +???? sin ?? =?? 2
This chord touches the circle ?? 2
+?? 2
=?? 2
,
Hence, Radius = Perpendicular distance of chord from centre.
?? =
?? 2
?? vcos
2
?? +sin
2
?? ??? 2
=???? .Hence ?? ,?? ,?? are in G.P .
Q10. A foot of the normal from the point (?? ,?? ) to a circle is (?? ,?? ) and a diameter of the circle has
the equation ?? ?? -?? =?? . Then the equation of the circle is
(a) ?? ?? +?? ?? +?? ?? -?? =??
(b) ?? ?? +?? ?? -?? ?? -?? =??
(c) ?? ?? +?? ?? -?? ?? -?? =??
(d) None of these.
Ans : (b) The line joining (4,3) and (2,1) is also along a diameter. So, the centre is the intersection of
the diameters 2?? -?? =2 and ?? -3=(?? -4). Solving these, the centre =(1,0)
? Radius = Distance between (1,0) and (2,1)=v2.
? Equation of circle (?? -1)
2
+?? 2
=(v2)
2
??? 2
+?? 2
-2?? -1=0
Q11. The polar of the point (?? ,-
?? ?? ) with respect to circle (?? -?? )
?? +?? ?? =?? is
(a) ?? ?? -???? ?? +?? =??
(b) ?? ?? -?? -???? =??
(c) ???? ?? -?? -???? =??
(d) ?? -???? ?? -?? =??
Ans : (b)
The polar of the point (5,-
1
2
) is ?? ?? 1
+?? ?? 1
+?? (?? +?? 1
)+?? (?? +?? 1
)+?? =0
?5?? -
1
2
?? -2(?? +5)+0+0=0?3?? -
?? 2
-10=0?6?? -?? -20=0.
Q12. The equation of the circle which touches the circle ?? ?? +?? ?? -?? ?? +?? ?? +???? =?? externally
and to which the lines ?? ?? -?? ???? -?? ?? +?? ?? =?? are normals, is
(a) ?? ?? +?? ?? -?? ?? -?? ?? -?? =??
(b) ?? ?? +?? ?? +?? ?? +?? ?? +?? =??
(c) ?? ?? +?? ?? -?? ?? -?? ?? +?? =??
(d) ?? ?? +?? ?? -?? ?? -?? ?? +?? =??
Ans : (d) Joint equations of normals are ?? 2
-3???? -3?? +9?? =0??? (?? -3?? )-3(?? -3?? )=0?
(?? -3)(?? -3?? )=0
? Given normals are ?? -3=0 and ?? -3?? =0, which intersect at centre of circle whose
coordinates are (3,1).
The given circle is ?? 1
=(3,-3),?? 1
=1;?? 2
=(3,1),?? 2
= ?
If the two circles touch externally, then ?? 1
?? 2
=?? 1
+?? 2
?4=1+?? 2
??? 2
=3
? Equation of required circle is (?? -3)
2
+(?? -1)
2
=(3)
2
??? 2
+?? 2
-6?? -2?? +1=0
Q13. There are two circles whose equations are ?? ?? +?? ?? =?? and ?? ?? +?? ?? -?? ?? -?? ?? +?? ?? =
?? ,?? ??? . If the two circles have exactly two common tangents, then the number of possible values
of ?? is
(a) 2
(b) 8
(c) 9
(d) None of these
Ans : (c) For ?? 2
+?? 2
=9, the centre =(0,0) and the radius =3
For ?? 2
+?? 2
-8?? -6?? +?? 2
=0. The centre =(4,3) and the radius =v(4)
2
+(3)
2
-?? 2
?4
2
+3
2
-?? 2
>0 or ?? 2
<5
2
or -5<?? <5.
Circles should cut to have exactly two common tangents.
So, ?? 1
+?? 2
>?? 1
?? 2
,?3+v25-?? 2
>v(4)
2
+(3)
2
or v25-?? 2
>2 or 25-?? 2
>4
??? 2
<21 or -v21<?? <v21
Therefore, common values of ?? should satisfy -v21<?? <v21.
But ?? ??? , So, ?? =-4,-3,……..3,4. ? Number of possible values of ?? =9.
Q14. If the circle ?? ?? +?? ?? =?? bisects the circumference of the circle ?? ?? +?? ?? -?? ?? +?? ?? +?? =?? ,
then ?? equals
(a) 4
(b) -4
(c) 16
(d) -16
Ans : (c) The common chord of given circles is ?? 1
-?? 2
=0?2?? -6?? -4-?? =0
Since, ?? 2
+?? 2
=4 bisects the circumferences of the circle ?? 2
+?? 2
-2?? +6?? +?? =0, therefore (i)
passes through the centre of second circle i.e. (1,-3). ? 2+18-4-?? =0??? =16.
Q15. A circle passes through the origin and has its centre on ?? =?? . If it cuts ?? ?? +?? ?? -?? ?? -?? ?? +
???? =?? orthogonally, then the equation of the circle is
(a) ?? ?? +?? ?? -?? -?? =??
(b) ?? ?? +?? ?? -?? ?? -?? ?? =??
(c) ?? ?? +?? ?? -?? ?? -?? ?? =??
(d) ?? ?? +?? ?? +?? ?? +?? ?? =??
Ans : (c) Let the required circle be ?? 2
+?? 2
+2???? +2???? +?? =0
This passes through (0,0), therefore ?? =0
The centre (-?? ,-?? ) of (i) lies on ?? =?? , hence ?? =?? .
Read More
|
209 videos|443 docs|143 tests
|
FAQs on Circle Solved Examples - Mathematics (Maths) for JEE Main & Advanced
| 1. How do you find the circumference of a circle? |  |
Ans. The circumference of a circle can be found using the formula: Circumference = 2 * π * radius.
| 2. What is the relationship between the circumference and diameter of a circle? | |
Ans. The circumference of a circle is directly proportional to its diameter, with the constant of proportionality being π (pi).
| 3. How is the area of a circle calculated? | |
Ans. The area of a circle can be calculated using the formula: Area = π * radius^2.
| 4. Can you find the radius of a circle if you know its circumference? | |
Ans. Yes, the radius of a circle can be found using the formula: radius = circumference / (2 * π).
| 5. How are the concepts of circles used in real-life applications? | |
Ans. Circles are used in various real-life applications such as designing wheels, gears, pulleys, and roundabouts.
Related Exams
About this Document
|
4.77/5 Rating |
|
Dec 26, 2024 Last updated |
Document Description: Circle Solved Examples for JEE 2024 is part of Mathematics (Maths) for JEE Main & Advanced preparation.
The notes and questions for Circle Solved Examples have been prepared according to the JEE exam syllabus. Information about Circle Solved Examples covers topics
like and Circle Solved Examples Example, for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for Circle Solved Examples.
Introduction of Circle Solved Examples in English is available as part of our Mathematics (Maths) for JEE Main & Advanced
for JEE & Circle Solved Examples in Hindi for Mathematics (Maths) for JEE Main & Advanced course.
Download more important topics related with notes, lectures and mock test series for JEE
Exam by signing up for free. JEE: Circle Solved Examples | Mathematics (Maths) for JEE Main & Advanced
Description
Full syllabus notes, lecture & questions for Circle Solved Examples | Mathematics (Maths) for JEE Main & Advanced - JEE | Plus excerises question with solution to help you revise complete syllabus for Mathematics (Maths) for JEE Main & Advanced | Best notes, free PDF download
Information about Circle Solved Examples
In this doc you can find the meaning of Circle Solved Examples defined & explained in the simplest way possible. Besides explaining types of
Circle Solved Examples theory, EduRev gives you an ample number of questions to practice Circle Solved Examples tests, examples and also practice JEE
tests
|
209 videos|443 docs|143 tests
|
Download as PDF
|
Explore Courses for JEE exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
Related Searches
Objective type Questions
,practice quizzes
,video lectures
,Previous Year Questions with Solutions
,past year papers
,ppt
,MCQs
,Sample Paper
,Semester Notes
,Important questions
,Extra Questions
,Exam
,Summary
,study material
,Circle Solved Examples | Mathematics (Maths) for JEE Main & Advanced
,Viva Questions
,shortcuts and tricks
,Circle Solved Examples | Mathematics (Maths) for JEE Main & Advanced
,mock tests for examination
,Free
,Circle Solved Examples | Mathematics (Maths) for JEE Main & Advanced
;
Additional Information about Circle Solved Examples for JEE Preparation
Circle Solved Examples Free PDF Download
The Circle Solved Examples is an invaluable resource that delves deep into the core of the JEE exam.
These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective.
With the help of these notes, you can grasp complex subjects quickly, revise important points easily,
and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner,
allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material,
or toppers' notes, this PDF has got you covered. Download the Circle Solved Examples now and kickstart your journey towards success in the JEE exam.
Importance of Circle Solved Examples
The importance of Circle Solved Examples cannot be overstated, especially for JEE aspirants.
This document holds the key to success in the JEE exam.
It offers a detailed understanding of the concept, providing invaluable insights into the topic.
By knowing the concepts well in advance, students can plan their preparation effectively.
Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.
Circle Solved Examples Notes
Circle Solved Examples Notes offer in-depth insights into the specific topic to help you master it with ease.
This comprehensive document covers all aspects related to Circle Solved Examples.
It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation.
Practice papers and question papers enable you to assess your progress effectively.
Additionally, the paper analysis provides valuable tips for tackling the exam strategically.
Access to Toppers' notes gives you an edge in understanding complex concepts.
Whether you're a beginner or aiming for advanced proficiency, Circle Solved Examples Notes on EduRev are your ultimate resource for success.
Circle Solved Examples JEE Questions
The "Circle Solved Examples JEE Questions" guide is a valuable resource for all aspiring students preparing for the
JEE exam. It focuses on providing a wide range of practice questions to help students gauge
their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation.
The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level.
Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.
Study Circle Solved Examples on the App
Students of JEE can study Circle Solved Examples alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the Circle Solved Examples,
students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc.
The EduRev App will make your learning easier as you can access it from anywhere you want.
The content of Circle Solved Examples is prepared as per the latest JEE syllabus.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup